Proposition 2

Let $A$ be a noetherian ring and $M$ a finitely generated Macaulay $A$-module. a) The $A$-module $M$ has no embedded associated prime ideals.\footnote{Recall (cf. IV, § 2, No. 3, remark) that an associated prime ideal of $M$ is said to be embedded if it is not a minimal element of $\mathrm{Supp}(M)$. Thus to say that $M$ has no embedded associated prime ideals means that the associated prime ideals of $M$ are the minimal elements of $\mathrm{Supp}(M)$.} b) Let $X$ be an irreducible closed subset of $\mathrm{Supp}(M)$ and $Y$ a closed subset of $X$. Then

c) The topological space $\mathrm{Supp}(M)$ is catenary (VIII, § 1, No. 2, def. 4).

d) Let $X_1$ and $X_2$ be irreducible components of $\mathrm{Supp}(M)$ and $Y$ a closed subset of $X_1 \cap X_2$. Then $\mathrm{codim}(Y, X_1) = \mathrm{codim}(Y, X_2)$.

a) Let $\mathfrak{p} \in \mathrm{Ass}(M)$. Then $\mathrm{prof}_A(\mathfrak{p}; M) = 0$ (§ 1, No. 1, remark 2), hence $\dim(M_{\mathfrak{p}}) = 0$ (No. 1, cor. of prop. 1), which implies that $\mathfrak{p}$ is a minimal element of $\mathrm{Supp}(M)$.

b) Suppose first that $Y$ is irreducible. Let $\mathfrak{p}$ and $\mathfrak{q}$ be the prime ideals of $\mathrm{Supp}(M)$ such that one has $Y = V(\mathfrak{q})$ and $X = V(\mathfrak{p})$. It follows from prop. 1 that one has

The general case follows from VIII, $\S$ 1, No. 2, remark 3.

c) Let $X$, $Y$, $Z$ be irreducible closed subsets of $\mathrm{Supp}(M)$ such that $Z \subset Y \subset X$. The codimension of each of these subsets in $\mathrm{Supp}(M)$ is finite (VIII, $\S$ 1, No. 4, prop. 9 and $\S$ 3, No. 1, cor. 1 of prop. 2). One then deduces from b) the equality

which implies c) (VIII, $\S$ 1, No. 2, prop. 4).

d) By b), we have $\mathrm{codim}(Y, X_1) = \mathrm{codim}(Y, \mathrm{Supp}(M)) = \mathrm{codim}(Y, X_2)$.

In particular, if there exists a finitely generated Macaulay $A$-module $M$, with support equal to $\mathrm{Spec}(A)$, the ring $A$ is catenary and consequently every ring of fractions or every quotient ring of $A$ is catenary (VIII, $\S$ 1, No. 3, remark 2).